**Class 10 Mathematics Quadratic Equations Extra Questions**

**Chapter 4: Quadratic Equations**

**Topic–1: Quadratic Equations **

**Very Short Answer Type Questions: [1 Mark] **

**1. If 8 is the root of the equation x**^{2}** – 10x + k = 0, then find the value of k. **

**Sol. **(8)2 – 10 × 8 + k = 0

⇒ 64 – 80 = –k ⇒ k = 16

**2. Is x = 3 a root of the equation 2x**^{2}** – 5x – 3 = 0? **

**Sol. **2(3)^{2} – 5(3) – 3 = 0

⇒ 18 – 15 – 3 = 0

So, x = 3 is the root of 2x^{2} – 5x – 3 = 0.

**3. (x + 2)**^{3}** = x(x**^{2}** – 1) is a quadratic equation. Is it true? **

**Sol. **(x + 2)^{3} =
x(x^{2} –
1)

⇒
x^{3} + 12x + 6x^{2} + 8 = x^{3} –
x

⇒
6x^{2} + 13x + 8 = 0

So, it is a quadratic equation.

**4. Is x = 1 a root of the equation 3x**^{2}** – 2x – 1 = 0? **

**Sol. **3(1)^{2} – 2 × 1 – 1 = 3 – 2 – 1 = 3 – 3 =
0

So, x = 1 is a root of 3x^{2} – 2x – 1 = 0

**5. For what value of k, x = 2 is a solution of kx**^{2}** + 2x – 3 = 0? **

Sol. k(2)^{2} + 2 × 2 – 3 = 0

⇒ 4k + 4 – 3 = 0

k = − ¼

**6. **If x = 2 is a root of the equation 3×2 – 2kx + 5 = 0, then find k.

**Sol.
**3 × (2)^{2} – 2 × 2 × k + 5 = 0

⇒ 12 – 4k + 5 = 0 ⇒ k = 17/4

** Short Answer Type Questions:**

**Which of the following among the given expressions is a polynomial?**

i) 1 + √x + x (ii) 1 + x^{2
}(iii) x + 1/x

**Sol. **(i) The power of x in √x, that is in x ^{1/2} is not a
non-negative integer.

∴ 1 + √x + x are not a polynomial.

(ii) Here, the power of x in every term of the given expression is non-negative.

∴ 1 + x^{2}
is a polynomial.

(iii) The power of x in the term 1/x, that is in, x^{–1} is a negative integer ∴ x + 1/x is not a polynomial.

Ans.

is a polynomial; since the degree of x in each term of the given expression is a non-negative integer. The degree of this polynomial is 2 since the maximum power of x in the given expression is 2.

is not a polynomial, since in 1/x^{2}, that is in x^{–2}, the power of x is a
negative integer.

** Short Answer Type questions: [3 Marks] **

**1. Check whether the following equations are quadratic or not : **

(i) (x – 1)(x – 3) = (x + 5)(x – 1)

(ii) x^{2} + 2x – 6 √x + 5 = 0

(iii) x^{3} – 4x^{2} + 1 = (x – 2)^{3}

**Sol.
**(i) (x – 1) (x – 3) = (x + 5)
(x – 1)

⇒ x^{2} – 4x + 3 = x^{2} + 4x –
5

⇒ –4x – 4x + 3 + 5 = 0

⇒ –8x + 8 = 0

Which is not of the
form ax^{2} + bx + c = 0.

Thus, (x – 1) (x – 3) = (x + 5) (x – 1) is not a quadratic equation.

(ii) x^{2} + 2x –
6 √x + 5 = 0

The given equation
contains a term involving √x, i.e., x ^{1/2}.

Since, x2 + 2x – 6 √x + 5 is not a quadratic polynomial, therefore,

x^{2} + 2x – 6 5 x + = 0 is not a quadratic equation.

(iii) x^{3} – 4x^{2}
+ 1 = (x – 2)^{3}

⇒ x^{3} – 4x^{2} + 1 = x^{3} – 8 – 6x(x – 2)

⇒ –4x^{2} +
1 = –8 – 6x^{2} + 12x

⇒ –4x^{2} +
6x^{2} – 12x + 1 + 8 = 0

⇒ 2x^{2} –
12x + 9 = 0, which is a
quadratic equation as it is of the form

ax^{2} + bx + c = 0. Thus, the given equation is a quadratic equation.

**2. **Determine whether the given values are solutions (roots) of the equation or not.

(ii) a^{2}x^{2} – 3abx + 2b^{2}= 0, x = a/b, x=b/a

⇒ 2 + 2 – 4 = 0

⇒ 4 – 4 = 0

⇒ 0 = 0, which is true.

x = √2 is a solution of the equation.

Now, putting x = −2 √2, we get

⇒ 8 – 4 – 4 = 0

⇒ 0 = 0, which is true.

So, x = −2 √2 is a solution of the equation.

(ii) a^{2}x^{2} – 3abx +
2b^{2} = 0

Putting x = a/b in the given equation, we get

⇒ a4 – 3a^{2}b^{2} + 2b^{4}
= 0, which is not true.

x = a/b is not a solution to the given equation.

Again, by putting x = b/a in the given equation, we get

⇒ b^{2} – 3b^{2}
+ 2b^{2} = 0

⇒ –2b^{2}
+ 2b^{2} = 0

⇒ 0 = 0, which is true.

x = b/a is a solution of the given equation.

the following, find the value of k for which the given value is a solution of the given equation :

(i) x^{2} – x(a + b) + k = 0, x = a

(ii) kx^{2} + √2x
– 4 = 0, x = 2

**Sol.
**(i) Since x = a is a root of the given equation. Therefore, it satisfies the
equation.

i.e., a^{2} – a (a + b) + k = 0

⇒ a^{2} – a^{2} – ab + k = 0

⇒ k = ab (ii) Since x = √2 is a root of the given equation. Therefore, it satisfies the equation.

⇒ b^{2} – 3b^{2}
+ 2b^{2} = 0

⇒ –2b^{2}
+ 2b^{2} = 0

⇒ 0 = 0, which is true.

x = b/a is a solution of the given equation.

**3. **In each of the following, find the value of k for which the given value
is a solution of the given equation :

(i) x^{2} – x(a + b) + k = 0, x = a

(ii) kx^{2} + √2x
– 4 = 0, x = 2

**Sol.
**(i) Since x = a is a root of the given equation. Therefore, it satisfies the
equation.

i.e., a^{2} – a (a + b) + k = 0

⇒ a^{2} – a^{2} – ab + k = 0

⇒ k = ab

(ii) Since x = √2 is a root of the given equation. Therefore, it satisfies the equation.

⇒ 2k + 2 – 4 = 0

⇒ 2k – 2 = 0 ⇒ k = 1

** Long Answer Type questions: [4 Marks] **

**1. If x = – 2 and x = − 1/5 are solutions of the equation 5x**^{2}** + kx + λ = 0, find the values of k and λ. **

**Sol. **Since x
= –2 and x = − 1/5 are solutions of the equation, 5×2 + kx + λ = 0

∴ 5(–2)^{2} + k(–2)
+ λ = 0

⇒ 2k – λ = 20 …(i)

From (i), we get

λ = 2k − 20

Substituting this value of λ in (ii), we get

k − 5(2k − 20) = 1

⇒ k −10k +100 = 1

⇒ –9k = –99 ⇒ k = 11

∴ λ = 2k – 20 = 2(11) – 20 = 2

Hence, k = 11 and λ = 2

**2. **If x = 2/3 and **x **= –3 are the roots of the equation a**x**2 + 7**x **+ b = 0, find the values of a and b. **Sol. **Since x = 2/3 is a root of ax2 + 7x + b = 0, we have,

⇒ 4a + 42 + 9b = 0

⇒ 4a + 9b + 42 = 0 …(i)

Again, x = –3
being a root of ax^{2}
+ 7x + b = 0, we have, a(–3)^{2}
+ 7(–3) + b = 0

⇒ 9a – 21 + b = 0

⇒ 9a + b – 21 = 0 …(ii)

Solving (i) and (ii), we get

a = 3 and b = – 6

**3. **If
a and b are the roots of the quadratic equation **x**2 – p ( **x **+ 1) –
c = 0, then find the value of (a + 1) (b + 1).

**Sol.
**We have x^{2}
– p(x + 1) – c = 0

⇒ x^{2} – px – (p + c) = 0

⇒ Since a and b are roots of the equation